If motional emf and induced emf are the same, how to explain the emf induced in a conducting rod moving with a velocity perpendicular to length of rod and magnetic field(magnetic field is perpendicular to length and velocity) in terms of induced electric field? There is no specific area present here, so magnetic flux is not varying so how does induced electric field come into play?

  • $\begingroup$ The area is the one that is being swept out by the rod. $\endgroup$
    – CuriousOne
    Commented Feb 25, 2016 at 10:55
  • $\begingroup$ @CuriousOne This is exactly the thing that I do not understand. Faraday's law states that the induced electromotive force in any closed circuit is equal to the rate of change of the magnetic flux enclosed by the circuit. Are you saying that the initial position of the rod finishes up as being the remote part of a loop the other sides being the current position of the rod and the trajectories of the ends of the rod? $\endgroup$
    – Farcher
    Commented Feb 25, 2016 at 11:24
  • $\begingroup$ @Farcher: Consider the electrical generator that generates the electricity for your computer: it's a long wire going round and round in a magnetic field! That wire connects to two brushes, which connect to a pair of long wires called transmission lines that connect to your wall plug and the power supply in your computer, which completes the loop (I left out a bunch of transformers for the fun of it). All of this is just a wire sweeping a magnetic field! Open up the loop... the electric field is still there. Leave the wires away... the electric field is still there! Relativity at work. :-) $\endgroup$
    – CuriousOne
    Commented Feb 25, 2016 at 11:33
  • $\begingroup$ @CuriousOne The electric field you mention is a non-conservative one? $\endgroup$
    – Farcher
    Commented Feb 25, 2016 at 12:01
  • $\begingroup$ @Farcher: Sure... if you can manage to crawl with a wire trough a generator running at a couple hundred rpm... why not. If you can't... then you simply can't tell the difference. $\endgroup$
    – CuriousOne
    Commented Feb 25, 2016 at 12:31

3 Answers 3


Motional emf and induced electric emf are not the same.

A motional EMF is due to magnetic forces and an induced electric EMF is due to electric forces.

And a motional EMF is only sometimes equal to the change in magnetic flux. It requires the wires to be thin and it requires the wires to keep the charges in the wire.

As for a rod moving perpendicular to itself and a constant uniform magnetic field the moving charges feel a magnetic force. The mobile charges are free to move they get the EMF, and the nonmobile charges get strained until the stress from the strain counters the magnetic force.

There is no specific area present here, so magnetic flux is not varying so how does induced electric field come into play?

There isn't an induced electric field. An EMF is not defined as $\int \vec E\cdot \mathrm d \vec \ell$ it is defined as the line integral of the force per unit charge, an example would be $\int \left(\vec E+\vec v\times \vec B\right)\cdot \mathrm d \vec \ell$ and when the loop is stationary then the velocity of the charges is in the same direction as $\mathrm d \vec \ell$ so the magnetic force doesn't contribute. When the loop is moving and the charge moves within the thin wire they can contribute.

For instance in your example the $\vec B$ field could be in the $\hat z$ direction, the velocity could be in the $\hat x$ dircetion and the wire could go in the direction $\mathrm d\vec \ell=\mathrm d \ell \hat y.$

For experts:

can I ask you how Faraday's Law is applied in this situation? Or put another way where is the loop through which the magnetic flux changes?

If you want to apply the universal flux rule $\left.\mathscr E\right|_{t=t_0}=$ $$\oint_{C(t_0)}\left(\vec E+\vec v\times \vec B\right)\cdot \mathrm d \vec \ell=-\left.\frac{\mathrm d \Phi_B}{\mathrm d t}\right|_{t=t_0}$$ then first you need to be aware it only holds when the wires are thin and the charges stay in the wire. Secondly you can pick any loop you want, but you do have to pick it for every moment of time to compute the flux through it as a function of time.

In particular if you care about the EMF along a bar then your area at any moment could be the set of places the bar has been since $t=0$ so the loop around the area can include the initial position if the bar as one part, the current position of the bar as another part and the locations the ends of the bar have been (between now and then) as the connecting parts. This loop can be used as the instantaneous loop $C(t_0)$ that encloses a surface $S(t_0)$ in the flux integral the line integral $\Phi_B(t)=\iint_{S(t)}\left(\vec B(t)\right)\cdot \hat n \mathrm d S$ for the changing flux. So the area swept out by the bar from $t=0$ to the current time can be the area you use to compute the function of time $\Phi_B(t).$

But the point of that choice is that the magnetic EMF only exists along the bar so the rest didn't matter. You could also just compute the Lorentz Force on the bar per unit charge in the instantaneous direction along the bar. It's the same thing. That's what an EMF is.

And sure, any loop based in time can have two contributions to the force per unit charge. One is the electric field. And around a closed loop that equals the flux of $\frac{\partial B}{\partial t}$ through the instantaneous position of the loop. The other is based on the magnetic force per unit charge. And for a thin wire where the motion is constrained by electric forces to stay in the wire over time, the magnetic for per unit charge.

If you want to think of it as the product rule you could compute $$\Phi_B(t+\Delta t)=\iint_{S(t+\Delta t)}\left(\vec B(t+\Delta t)\right)\cdot \hat n(t+\Delta t)\mathrm d S(t+\Delta t)$$

and $$\Phi_B(t)=\iint_{S(t)}\left(\vec B(t)\right)\cdot \hat n(t)\mathrm d S(t).$$

So when you subtract and divide by $\Delta t$ you can instead look at $$\iint_{S(t)}\left(\frac{\vec B(t+\Delta t)-\vec B(t)}{\Delta t}\right)\cdot \hat n(t)\mathrm d S(t)$$ and $$\frac{1}{\Delta t}\iint_{S(t+\Delta t)}\left(\vec B(t)\right)\cdot\hat n(t+\Delta t) \mathrm d S(t+\Delta t)-\frac{1}{\Delta t} \iint_{S(t)}\left(\vec B(t)\right)\cdot\hat n(t) \mathrm d S(t).$$

And the limit of their sum equals $\mathrm d \Phi_B/\mathrm d t$ when the wires are thin, there are no magnetic monopoles, and the charges stay in the wire.

  • $\begingroup$ I think I am slowly getting there but initially with the non experts. . When a conducting rod moves through a uniform B-field there is a force on a charge $F_B=q\vec v \times \vec B$. So charges accumulate at the ends and the separation of charges sets up an electric field such that the net force on charges in the rod is zero $F_B = F_e \Rightarrow Bqv = qE$. There is a potential difference across the rod. That potential difference is numerically equal to the motional emf. $\endgroup$
    – Farcher
    Commented Feb 26, 2016 at 16:45
  • $\begingroup$ @Farcher Scalar potential is a gauge dependent quantity, it's not objective in any way and it's not what is measured by a voltmeter. A voltmeter measures an EMF around the loop that includes the voltmeter. An EMF is the force per unit charge line integrated around a loop, and it depends on the path of the loop (unlike a voltage difference which doesn't depend on the path). The motional EMF exists the instant the conductor begins to move. A charge buildup comes later and if it's an isolated conductor the equilibrium has zero line integral of force per unit charge between the ends of the rod $\endgroup$
    – Timaeus
    Commented Feb 26, 2016 at 16:57
  • $\begingroup$ I understand the build up of charges after the motional emf instantaneous birth. So I wait until I get a steady state and there is an imbalance of charges at each end. No further charges are accumulating at the end. What keeps the mobile charges in the middle of the rod from moving? What force opposes $F_B$? $\endgroup$
    – Farcher
    Commented Feb 26, 2016 at 17:46
  • $\begingroup$ @Farcher After the isolated bar reaches equilibrium then there is an electric force equal and opposite to the magnetic force along the bar. Of course there is now an additional magnetic field besides the original external one, this one from the moving charge imbalance on the ends. But if it's all symmetrical the magnetic field from the moving charge imbalance can be zero on the center of the bar where the magnetic and electric forces cancel. $\endgroup$
    – Timaeus
    Commented Feb 26, 2016 at 17:53
  • $\begingroup$ I forgot about the moving charges at the end producing a magnetic field. But relative to the charges in the wire that magnetic field is not moving? So does it have a effect? Now that we have an electric force in play must that not come from the electric field produced by the separation of charges? $\endgroup$
    – Farcher
    Commented Feb 26, 2016 at 18:45

enter image description here

Imagine a loop $ABCC'B'A'$ which is in a magnetic field $\vec B$ as in the left-hand diagram.
Assume that sides $AA'$ and $CC'$ are close together and the magnetic flux through the loop is some value $\Phi_i$.
Suppose that this is an instantaneous picture of the loop at time $t_i$ and that side $CC'$ is moving at constant velocity $\vec v$ in a direction which is at right angle to the magnetic field and the side $CC'$.
The right-hand diagram shows side $CC'$ at a time $t_f$ and the magnetic flux through that loop is now $\Phi_f$.

There is change of magnetic flux through the loop and so Faraday says that there is an emf induced in the loop, $\frac{\Phi_f-\phi_i}{t_f - t_i}$.

If $CC'$ was a conducting rod all that was stated above would still be the same.

To illustrate a point, in the left-hand diagram I have portions of the loop hanging down.
Those portions could equally well be elastic joining $A$ and $C$ and $A'$ and $C'$ directly which were stretched when side $CC'$ moved.

Or you could save yourself a lot of trouble and use "The area is the one that is being swept out by the rod" as was stated by @CuriousOne noting that the area must be at right angles to the magnetic field direction $\int_A \vec B \cdot d\vec a$.

  • $\begingroup$ It is not just the area swept by the rod. It is the area swept by the rod the "cut" through magnetic field lines. If your sweeping surface moves parallel to the field lines (for a field supposed constant in time) without cutting any, you won't have any emf (still by applying Faraday's law) for your flux would not have changed. $\endgroup$
    – A.G.
    Commented Feb 25, 2016 at 15:17
  • $\begingroup$ @A.G. I agree, then there would be no change of magnetic flux so I have added an extra few words to my last sentence. $\endgroup$
    – Farcher
    Commented Feb 25, 2016 at 15:47
  • $\begingroup$ This is fundamentally flawed. If your magnetic field is unchanging in time, then the electric field is conservative, it could even be zero. Faraday can get an EMF equal to the change in magnetic flux, if the loops are thin, if the charges stay in the wire, and even so the EMF will not be due to electric fields. $\endgroup$
    – Timaeus
    Commented Feb 25, 2016 at 18:03
  • $\begingroup$ @Timaeus I have read carefully your comment but now I am unsure about the loop through which the magnetic flux is changing in which the emf is generated. The end result of the rod moving through the magnetic field is an accumulation of charges at the ends of the rod? How is these charges do not produce an electric field? $\endgroup$
    – Farcher
    Commented Feb 25, 2016 at 19:49
  • $\begingroup$ @Farcher Whether you get charges at the end or on the sides depends on whether the moving bar is connected to other conductors. If it is just an isolated bar that you pull at constant speed then you'll eventually develop charges on the end and at that point there will be no EMF because the electric and magnetic EMFs along the bar cancel. If it's connected to some stationary conductors to complete a loop and the bar has a finite resistance then instead charge will develop on the sides of the moving bar to keep the steady current inside the wire (a Hall voltage) and there will be a magnetic EMF $\endgroup$
    – Timaeus
    Commented Feb 25, 2016 at 19:54

Great and beautiful question ! Please patiently read through this ! I have tried my best to elucidate it !

The thing is you are talking about two different kinds of induced emfs here ! one is a conservative and the other non-conservative. If you think of the emf in the rod if you start from one end of the rod and move along the entire length the total emf change is non zero. but in the case of the loop present in the magnetic field, with reference to which you are talking about "area", the magnetic field change or area change that you create induces an emf but this emf is non conservative ! meaning if you move along the loop you will come back to the same point, and thus total emf must be zero ! but it isnt zero (or is it ?) ! Because there is a current ! Due to this fundamental difference of the two kinds of EMFs being completely different you are faced with this phenomena !

I would suggest you this. think of the EMF formation as a result of the electrons getting pushed by the varying magnetic field around them. thus if u see in both cases you have the electrons getting pushed and thus it becomes obvious that there must be emfs in both cases. If the magnetic flux is changing inside the loop then field lines must be either entering the loop or leaving it. thus this must happen by the field lines intersecting the wire at some point of entering or leaving ! I would also add that in the case of the rod the emf can manifest itself at the two ends ! Well because there are "ends" for it to manifest at ! But there is ideally no current ! This is a voltage that primarily appears as increased potential of one point compared to the other ! But in the loop there are no ends thus there is no measurable EMF ! but there is a current in a loop with has a finite inductance ! This is the other Manifestation of EMF ! This is another Voltage that manifests as an induced current against a finite resistance (impedance) !

  • 1
    $\begingroup$ This induced electric field exists without any charge carriers. It's a (relativistic) property of the electromagnetic field. $\endgroup$
    – CuriousOne
    Commented Feb 25, 2016 at 11:20
  • $\begingroup$ @CuriousOne There isn't an induced electric field because the OP never said the magnetic field changed in time (or in space). $\endgroup$
    – Timaeus
    Commented Feb 25, 2016 at 18:05
  • $\begingroup$ @Timaeus: There doesn't have to be. All that's needed is a constant magnetic field and movement. $\endgroup$
    – CuriousOne
    Commented Feb 25, 2016 at 20:14
  • $\begingroup$ @CuriousOne That was my point. When there is a constant uniform magnetic field and a moving wire there is an EMF from the magnetic force on the moving charges. There isn't an electric field for a uniform unchanging magnetic field. Not every EMF is caused by electric forces, a motional EMF is not caused by electric fields. There isn't an induced electric field, so any answer to the OP that doesn't tell them there isn't an induced electric field is not a correct answer. $\endgroup$
    – Timaeus
    Commented Feb 25, 2016 at 20:18

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